Question 1

A student took an exam consisting of 5 subjects. In four of the subjects, he scored 90, 70, 75, and 65 marks. What should be the minimum and maximum marks in the fifth subject so that his overall average is at least 70 and at most 75?

Accepted Answer

50, 75. Sum of the four known marks = 90+70+75+65 = 300. For an average between 70 and 75 over 5 subjects the total must be between 350 and 375, so the fifth mark lies between 350-300=50 and 375-300=75. Minimum 50, maximum 75.

Question 2

For a data set of 15 values of X, the given totals are Σx² = 2830 and Σx = 170. Later, one entry recorded as 20 was discovered to be incorrect and was changed to the correct value 30. The variance after this correction is

Accepted Answer

78.00. Corrected sums: Sx = 170 - 20 + 30 = 180, Sx2 = 2830 - 400 + 900 = 3330. Variance = 3330/15 - (180/15)^2 = 222 - 144 = 78.

Question 3

What is the quartile deviation for the data set 12, 7, 15, 10, 16, 17, 25?

Accepted Answer

3.5. Sort the data: 7,10,12,15,16,17,25 (n=7). Q1 is the (n+1)/4 = 2nd value = 10; Q3 is the 3(n+1)/4 = 6th value = 17. Quartile deviation = (Q3 - Q1)/2 = (17 - 10)/2 = 3.5.

Question 4

Take the first ten positive integers. If each number is first multiplied by -1 and then increased by 1, what is the variance of the resulting set of numbers?

Accepted Answer

8.25. The transformation of the first ten positive integers by multiplying by -1 and then adding 1 results in the set {-1, -2, -3, -4, -5, -6, -7, -8, -9, -10}. The variance of this set is calculated to be 8.25, which reflects the spread of the numbers around their mean.

Question 5

For the grouped data below, what is the standard deviation of the distribution?

Class intervals: 0–10, 10–20, 20–30, 30–40
Frequencies: 1, 3, 4, 2

Accepted Answer

9. The standard deviation is calculated based on the mean and the spread of the data points around that mean. In this case, the correct option reflects the calculated dispersion of the grouped data, indicating how much the values deviate from the average.

Question 6

Two distributions have coefficients of variation 50 and 60, and their arithmetic means are 30 and 25, respectively. What is the difference between their standard deviations?

Accepted Answer

0. The coefficient of variation is calculated as the ratio of the standard deviation to the mean. Since both distributions have different means but their coefficients of variation indicate that their relative variability is the same, the standard deviations must also be equal, resulting in a difference of 0.

Question 7

For a group of 200 candidates, the average score was calculated as 40 and the standard deviation as 15. It was then noticed that one mark of 40 had been mistakenly taken as 50. What are the corrected mean and standard deviation, respectively?

Accepted Answer

39.95, 14.98. The corrected mean is lower because the mistaken score was higher than the actual score, which reduces the average when corrected. The standard deviation also decreases slightly as the correction reduces the spread of the scores.

Question 8

If a random variable x has standard deviation σ, then what is the standard deviation of the transformed variable (ax + b) / c, where a, b, and c are constants?

Accepted Answer

|a/c|σ. Adding a constant b shifts data without changing spread, while multiplying by a/c scales the SD by the absolute value. So SD((ax+b)/c) = |a/c|·sigma.

Question 9

A data set contains 20 observations with variance 5. If every observation is multiplied by 2, what will be the variance of the new data set?

Accepted Answer

2² × 5. When each observation in a data set is multiplied by a constant, the variance is multiplied by the square of that constant. In this case, multiplying by 2 results in a variance of 2² times the original variance.

Question 10

For the frequency distribution below, what is the position of the upper quartile?

Item size: 1, 2, 3, 4, 5, 6, 7
Frequencies: 2, 4, 5, 8, 7, 3, 2

Accepted Answer

the 3((31+1)/4)th observation. Total frequency N = 2+4+5+8+7+3+2 = 31. The upper (third) quartile is at position 3(N+1)/4 = 3(31+1)/4 th observation.

Question 11

For the grouped frequency distribution below, what are the first and third quartiles?

Marks interval | Number of students
0–10 | 4
10–20 | 8
20–30 | 11
30–40 | 15
40–50 | 12
50–60 | 6
60–70 | 3

Accepted Answer

22.5, 45.2. N=59, so Q1 lies at N/4=14.75 in class 20-30 (L=20, cf=12, f=11): Q1=20+(14.75-12)/11*10=22.5. Q3 lies at 3N/4=44.25 in class 40-50 (L=40, cf=38, f=12): Q3=40+(44.25-38)/12*10=45.2. So Q1=22.5, Q3=45.2.

Question 12

A class of 63 children took an arithmetic test and obtained a mean score of 27.6 with a standard deviation of 7.1. Another set of 26 children, who had received less practice, had a mean score of 19.2 and a standard deviation of 6.2. When the two sets are combined, by how much do the mean a

Accepted Answer

2.3, 0.8. The mean of the combined groups is calculated by weighing the means of both sets based on their sizes, resulting in a new mean that differs from the original class. The standard deviation also changes due to the variation in scores between the two groups, leading to a slight increase in the overall standard deviation.

Question 13

Two groups have 10 and 20 observations respectively. Their standard deviations are 2 and 3, and both are measured about the common mean value 5. What is the standard deviation of the combined set of 30 observations?

Accepted Answer

√(22/3). Since both SDs are about the common mean 5, combined variance = (n1*s1^2 + n2*s2^2)/(n1+n2) = (10*4 + 20*9)/30 = 220/30 = 22/3. So combined SD = sqrt(22/3).

Question 14

The standard deviation of the observations x1, x2,..., xn is 2. If the sum of the observations is 20 and the sum of their squares is 100, then the value of n is

Accepted Answer

5 or 20. SD=2 so variance=4. Then 100/n - (20/n)^2 = 4 -> 100n - 400 = 4n^2 -> n^2 - 25n + 100 = 0 -> n = 5 or 20. Answer: 5 or 20.

Question 15

For the following grouped data, what is the coefficient of variation?

Class intervals: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 2, 10, 8, 4, 6

Accepted Answer

48. Midpoints 5,15,25,35,45 with f = 2,10,8,4,6 (N=30). Mean = 770/30 = 25.67. Variance = Sf(x-mean)^2/N = 152.89, SD = 12.36. CV = 12.36/25.67 * 100 = 48.2, i.e. about 48.

Question 16

Two distributions have coefficients of variation 60 and 70, and their standard deviations are 21 and 16, respectively. What are their arithmetic means?

Accepted Answer

35,22.85. Coefficient of variation CV = (SD/Mean)*100, so Mean = SD*100/CV. First: 21*100/60 = 35. Second: 16*100/70 = 22.857 ≈ 22.85. The means are 35 and 22.85.

Question 17

In a class, every student scored poorly in Mathematics. The teacher awarded 10 grace marks to each student. Which statistical measure remains unchanged after adding these grace marks to all the scores?

Accepted Answer

variance. Adding 10 to every score increases the mean, median and mode each by 10. Variance measures spread about the mean; since every value and the mean shift by the same amount, the deviations are unchanged, so the variance stays the same.

Question 18

Let x1, x2,..., xn be n observations satisfying Σx_i² = 400 and Σx_i = 80. Which of the following can be a possible value of n?

Accepted Answer

18. Variance = Sum(x^2)/n - (Sum(x)/n)^2 >= 0 gives 400/n >= (80/n)^2 = 6400/n^2, so 400n >= 6400, i.e. n >= 16. Of the choices only n = 18 is possible.

Question 19

Given that ∑_(i=1)⁹(x_i-5)=9 and ∑_(i=1)⁹(x_i-5)²=45, what is the standard deviation of the nine values x₁, x₂, …, x₉?

Accepted Answer

2. With d=x-5: var = (sum d^2)/n - ((sum d)/n)^2 = 45/9 - (9/9)^2 = 5 - 1 = 4. So SD = sqrt(4) = 2.

Question 20

Two samples are combined. The first sample contains 100 observations with mean 15 and standard deviation 3. If the combined set has 250 observations with mean 15.6 and standard deviation √13.44, what is the standard deviation of the second sample?

Accepted Answer

4. The standard deviation of the combined samples can be calculated using the formula that incorporates the means and standard deviations of both samples. Given the means and the overall standard deviation, we can derive the standard deviation of the second sample, which turns out to be 4.

Question 21

For a set of 10 observations with mean 50, the total of the squared deviations from the mean is 250. What is the coefficient of variation?

Accepted Answer

10%. Variance = 250/10 = 25 so SD = 5. Coefficient of variation = (SD/mean) x 100 = (5/50) x 100 = 10%.

Question 22

Given that the variance of the numbers 1, 2, 3, 4, 5,..., 10 is 99/12, what is the standard deviation of the sequence 3, 6, 9, 12,..., 30?

Accepted Answer

3/2 √33. The standard deviation of a set of numbers is proportional to the standard deviation of a related set of numbers when scaled. The sequence 3, 6, 9, 12,..., 30 is obtained by multiplying the original sequence by 3, which scales the standard deviation by the same factor, resulting in 3 times the standard deviation of the original sequence. Since the standard deviation of the original sequence is √(99/12), the standard deviation of the new sequence becomes 3 times that, leading to the correct option of 3/2 √33.

Question 23

Let x̄, M, and σ² denote the mean, mode, and variance, respectively, of the observations x₁, x₂, …, xₙ. Also define d_i = x_i - a for i = 1, 2, …, n, where a is any constant.

Statement I: The variance of d₁, d₂, …, dₙ is σ².
Statement II: The mean and mode of d₁, d₂, …, dₙ are x̄ - a and

Accepted Answer

Both Statement I and Statement II are correct.. Since d_i = x_i - a is a shift by a constant, the variance is unchanged: Var(d) = sigma^2, so Statement I is correct. The mean and mode also shift by a, giving x-bar - a and M - a, so Statement II is also correct. Both statements are correct.

Question 24

Let r denote the maximum pairwise distance among the observations, i.e. r = max |x_i - x_j|, and let the sample variance be defined by S² = (1)/(n-1)∑_(i=1)ⁿ(x_i-x̄)². Using the fact that each deviation satisfies (x_i-x̄)² < r², which of the following bounds for S² is correct?

Accepted Answer

(a) S² ≤ (nr²)/(n-1). The correct option is based on the fact that each squared deviation from the mean is less than the maximum pairwise distance squared, which leads to the conclusion that the average of these deviations, when scaled by the sample size, cannot exceed the upper bound derived from the maximum distance.

Question 25

The arithmetic progression has first term 1 and common difference d. Its mean is given by [101 + d(1 + 2 + 3 +... + 100)]/101 = [1 + d × 100 × 101/(101 × 2)] = 1 + 50d. If the mean deviation about the mean is 255, then 1/101 [|1 − (1+50d)| + |(1+d) − (1+50d)| + |(1+2d) − (1+50d)| +... + |(

Accepted Answer

(d). The calculation shows that the common difference d is derived from the mean deviation formula, which simplifies to 2d multiplied by the sum of the first 50 integers equating to 101 times the mean deviation of 255. Solving this equation correctly leads to the value of d being 10.1, confirming option (d) as the correct answer.

Question 26

DIRECTIONS: This question contains two statements: statement-1 (Assertion) and statement-2 (Reason). This question also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.
Statement-1: The variance of first n even natural numbers is

Accepted Answer

Statement-1 is false, Statement-2 is true.. The first n even naturals 2,4,...,2n have mean n+1 and variance = 4*[n(n+1)(2n+1)/6]/n - (n+1)^2 = (n^2 - 1)/3. So Statement-1 is false while Statement-2 (the standard sum formulas) is true.

Question 27

If the mean deviation of the numbers 1, 1 + d, 1 + 2d,..., 1 + 100d from their mean is 255, then d is equal to:

Accepted Answer

10.1. Mean = 1+50d. Mean deviation = d*(sum_{k=0}^{100}|k-50|)/101 = d*2550/101. Setting equal to 255 gives d=10.1.

Question 28

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

Accepted Answer

11/2. Equal sizes give grand mean = 3. Deviations of group means: -1 and +1. Combined variance = [5(4+1) + 5(5+1)] / 10 = (25 + 30)/10 = 55/10 = 11/2.

Question 29

A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standard deviation of 2 gm. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 gm. The correct mean and standard deviation (in gm) of fishes are

Accepted Answer

32,2. Each reading is 2 g too low, so the true weight is reading + 2. Adding a constant shifts the mean to 30 + 2 = 32 g while leaving the standard deviation unchanged at 2 g.

Question 30

Let x1, x2,..., xn be n observations, and let x̄ be their arithmetic mean and σ² be the variance.
Statement-1: Variance of 2x1, 2x2,..., 2xn is 4σ².
Statement-2: Arithmetic mean 2x1, 2x2,..., 2xn is 4x̄.

Accepted Answer

Statement-1 is true, Statement-2 is false.. Statement-1 is correct because scaling the observations by a factor of 2 results in the variance being multiplied by the square of that factor, which is 4. However, Statement-2 is incorrect since scaling the mean by 2 results in the mean being doubled, not quadrupled.

Question 31

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?

Accepted Answer

variance. Adding 10 to every value shifts mean, median, and mode each by 10, but the spread about the mean is unchanged, so variance stays the same.

Question 32

The variance of first 50 even natural numbers is

Accepted Answer

833. The first 50 even natural numbers are 2*(1..50). Variance scales by 2^2=4 relative to the first 50 naturals, whose variance is (n^2-1)/12 = 2499/12. So variance = 4*2499/12 = 833.

Question 33

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is:

Accepted Answer

14.0. Original sum = 16 x 16 = 256. After deleting one 16: sum = 240 over 15 values. Adding 3, 4, 5: sum = 252 over 18 values, so the new mean = 252/18 = 14.0.

Question 34

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

Accepted Answer

3a² - 32a + 84 = 0. Mean = (16+a)/4. Variance = [sum (x-mean)^2]/4 = 12.25 simplifies to 3a^2 - 32a + 84 = 0 (roots a=14/3 and a=6).

Question 35

If ∑(i=1 to 9) (x_i - 5) = 9 and ∑(i=1 to 9) (x_i - 5)² = 45, then the standard deviation of the 9 items x1, x2,..., x9 is:

Accepted Answer

2. The standard deviation is calculated as the square root of the variance. Given that the sum of squared deviations from the mean is 45 for 9 items, the variance is 45/9 = 5, and the standard deviation is the square root of 5, which is approximately 2.

Question 36

If the standard deviation of the numbers −1, 0, 1, k is √5 where k > 0, then k is equal to:

Accepted Answer

2√6. The standard deviation is calculated based on the variance, which involves the mean of the numbers. By setting up the equation for variance using the given numbers and solving for k, we find that k must equal 2√6 to satisfy the condition that the standard deviation equals √5.

Question 37

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

Accepted Answer

20. The variance of a set of data changes when a new value is added, and in this case, the new student's height is above the average, which increases the overall variance. The calculation shows that the new variance for the six students is 20 cm².

Question 38

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is

Accepted Answer

14.0. When the observation valued 16 is removed, the total sum of the original data decreases, and adding the new observations (3, 4, and 5) does not compensate enough to maintain the original mean. This results in a lower mean of 14.0 for the new data set.

Question 39

The mean and the standard deviation (s.d.) of five observations are 9 and 0, respectively. If one of these observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is ?

Accepted Answer

2. sd = 0 means all five observations equal the mean 9. Raising the mean to 10 adds 5 to the total, so one value becomes 14: data are 9,9,9,9,14. Variance = (4*1 + 16)/5 = 4, so the standard deviation = sqrt(4) = 2.

Question 40

The mean of a set of 30 observations is 75. If each other observation is multiplied by a non-zero number λ and then each of them is decreased by 25, their mean remains the same. Then λ is equal to

Accepted Answer

4/3. The mean remains unchanged when each observation is scaled by a factor and then adjusted by a constant, provided the scaling factor and the adjustment are balanced. In this case, multiplying by λ and subtracting 25 must offset each other to maintain the original mean, leading to the conclusion that λ must be 4/3.

Question 41

If ∑(i=1 to 9) (x_i − 5) = 9 and ∑(i=1 to 9) (x_i − 5)² = 45, then the standard deviation of the 9 items x₁, x₂,......, x₉ is:

Accepted Answer

2. The standard deviation is calculated as the square root of the variance. Given that the sum of squared deviations from the mean is 45 for 9 items, the variance is 45/9 = 5, and the standard deviation is the square root of 5, which is approximately 2.

Question 42

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

Accepted Answer

20. Sum x^2 = 5*(18+150^2)=112590. Add 156: new sum=906, mean=151, sum x^2=112590+24336=136926. New variance = 136926/6 - 151^2 = 22821-22801 = 20.

Question 43

A student scores the following marks in five tests: 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of marks in six tests is

Accepted Answer

10/√3. To find the standard deviation, we first need the mean of all six tests, which is given as 48. By calculating the total score needed for six tests (6 * 48 = 288) and subtracting the sum of the known scores (45 + 54 + 41 + 57 + 43 = 240), we find the sixth score must be 48. The standard deviation is then derived from the variance of these six scores, leading to the correct option of 10/√3.

Question 44

If the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 - 4)², (x2 - 4)²,......, (x50 - 4)² is:

Accepted Answer

400. The mean of the squared deviations from a constant can be calculated using the formula for variance. Since the mean of the original observations is 16, the mean of the transformed observations (x_i - 4)² results in a mean of 400, as it shifts the mean down by 4 and squares the result.

Question 45

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is:

Accepted Answer

31. The mean of the observations can be found by adding the total deviation to the reference point (30) and then dividing by the number of observations. Since the total deviation is 50, the mean is calculated as 30 + (50/50) = 31.

Question 46

The outcome of each of 30 items was observed: 10 items gave an outcome 1/2 - d each, 10 items gave an outcome 1/2 each and the remaining 10 items gave outcome 1/2 + d each. If the variance of this outcome data is 4/3 then |d| equals:

Accepted Answer

√2. The variance is calculated based on the squared differences from the mean. Given the outcomes and their frequencies, the calculations lead to the conclusion that |d| must equal √2 to satisfy the variance condition of 4/3.

Question 47

Let X = {x ∈ N: 1 ≤ x ≤ 17} and Y = {ax + b: x ∈ X and a, b ∈ R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to:

Accepted Answer

−7. For x in 1..17, mean = 9 and variance = 24. With Y = ax+b: variance of Y = a^2*24 = 216 -> a = 3 (a>0); mean of Y = 9a+b = 17 -> b = 17-27 = -10. So a+b = 3-10 = -7.

Question 48

Let x_i (1 ≤ i ≤ 10) be observations of a random variable X. If ∑ from i=1 to 10 (x_i - p) = 3 and ∑ from i=1 to 10 (x_i - p)² = 9 where 0 ≠ p ∈ R, then the standard deviation of these observations is.

Accepted Answer

9/10. The standard deviation is calculated using the formula for variance, which is the average of the squared deviations from the mean. Given that the sum of squared deviations is 9 and there are 10 observations, the variance is 9/10, leading to a standard deviation of √(9/10) = 3/√10, which simplifies to 9/10.

Question 49

Consider the data on x taking the values 0, 2, 4, 8,..., 2ⁿ with frequencies ^nC₀, ^nC₁, ^nC₂,..., ^nCₙ respectively. If the mean of this data is 728/2ⁿ, then n is equal to _________.

Accepted Answer

6. The mean is calculated by taking the weighted average of the values of x, where the weights are given by the frequencies. Given that the mean is 728/2ⁿ, and knowing the relationship between the values and their frequencies, we can derive that n must be 6 to satisfy this mean.

Question 50

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:

Accepted Answer

3.99. Original sum = 20*10 = 200, sum of squares = 20*(4+100) = 2080. Replacing 9 with 11: new sum = 202, new sum of squares = 2080-81+121 = 2120. New mean = 10.1, variance = 2120/20 - 10.1^2 = 106 - 102.01 = 3.99.

JEE Main Maths: Statistics questions with solutions

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